Step 1: Understand the statement.
We must simplify $(p\to q)\land(q\to\;\sim p)$ and match it to one of the options. The neat way is to make a truth table for all four cases of $p$ and $q$.
Step 2: Recall the rule for implication.
$A\to B$ is false only when $A$ is true and $B$ is false; in every other case it is true.
Step 3: Check the case $p$ true, $q$ true.
Then $p\to q$ is T. But $q\to\;\sim p$ is T$\to$F which is F. So the whole AND is F.
Step 4: Check the case $p$ true, $q$ false.
Then $p\to q$ is T$\to$F which is F. So the whole AND is F.
Step 5: Check the cases where $p$ is false.
If $p$ is false then $\sim p$ is true. With $p$ false, $p\to q$ is automatically T. Also $q\to\;\sim p$ ends in a true value, so it is T. So the AND is T in both $q$ true and $q$ false sub-cases.
Step 6: Read the pattern.
The compound statement is true exactly when $p$ is false, and false when $p$ is true. That is exactly the statement $\sim p$. \[ \boxed{(p\to q)\land(q\to\;\sim p)\equiv\;\sim p} \]